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2D Forces

  1. Oct 4, 2008 #1
    1. The problem statement, all variables and given/known data
    The person wants to cause the 200N crate to start sliding toward the right. To achieve this, the horizontal component of the force exerted on the crate by the rope must equal 0.35 times the normal force exerted on the crate by the floor.

    [[[[[[[[[ +
    [........[ +

    The box is the crate and the pluses represent the rope the guy is pulling. It makes an angle of 20 degrees with the horizontal.
    2. Relevant equations

    3. The attempt at a solution
    So I figured that 0.35 is the [tex]\mu[/tex] for friction. Since its a leveled surface, FN will equal Fg and thus they cancel out. The horizontal component of the Fapp has to equal Ff because he's trying to move it, not accelerate it.

    From this, I got Fapp to equal 70N. The angle is 20 degrees, so the Fapp is 74.5N (got by: 70N/cos(20)). But the back of the book is telling me it should be 66N. I don't know how I'd get that. But then again, its Engineering Statics, so maybe there is some trick? Please help?
  2. jcsd
  3. Oct 4, 2008 #2
    Well you have a problem here. You say the only y forces are normal force and gravity.

    However there is a y component of the man pulling the crate.
    You need to form two simultaneous equations.
    You do NOT know the mass of the crate.

    So do this: let x = the force he is pulling with (hypotenuse).

    [tex] 200 = mg + xsin(20) [/tex]
    [tex] xcos(20) = 0.35mg [/tex]

    Now explain to me why these equations are as they are.
    You will get the right answer.
  4. Oct 4, 2008 #3
    Why do we not know the mass of the crate? We know the weight, which is m*9.81?

    Okay.. I'll try:
    200=mg+xsin(20).. In this one 200 is the upward force created by two things - the man lifting the object (xsin(20) is the y-component of him lifting it) and mg is simply opposite the force the ground is exerting on the object upwards. They're added because they are the forces acting up.

    xcos(20)=0.35mg, well that's a given. The horizontal component of the applied force is 0.35 of the natural force.

    I can picture this in a FBD, I think. Thanks!
  5. Oct 4, 2008 #4
    we know the net force in the negative y direction, 200N. However, this is mg + xsin(20) because there is mg pushing down and a small component of the man pulling up. If he was only pulling horizontally, it would be m*9.81.
    We can however, using those simultaneous equations solve for the mass of the crate. When I said we do not know the mass, I meant, we have not been given the mass in a simple form.

    Your explanation is enirely right.
  6. Oct 4, 2008 #5
    What about this though?

    Its the same box, but I don't know how to set this up. Isn't the angle changing as he's pulling the rope upwards?
  7. Oct 5, 2008 #6
    yes, in that example it is. However, in the first example that you posed; he is pulling parallel to the angle of the chain. Therefore the angle of the chain is NOT changing but there is still an X and Y component of force.
  8. Oct 5, 2008 #7
    Yeah, I know in the first one it wasn't changing. How would I solve this one though?
  9. Oct 5, 2008 #8
    What are you given, and what is the question asking?
  10. Oct 5, 2008 #9
    It's the same question. Only now he anchored the cable to a point (the blue circle on the right) and is pulling the rope upwards. The 10 degrees refers to the angle the rope makes with the ground at the point where it's anchored. Same box (200N).
  11. Oct 5, 2008 #10


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    And the rope attached to the crate is horizontal?
  12. Oct 5, 2008 #11
    It's not parallel to the ground in the book... So I guess not.
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